How to prove that a subgroup of a group is normal based on generating sets? I apologize if this is a duplicate question, but I read online that one method by which to show that a subgroup is normal is by means of generating sets (if both groups have known presentations). In other words, you would simply show that conjugating the generators of the subgroup by generators of the main group results in only members of the subgroup. But is there a proof of this fact? If my statement is unclear I apologize again, I am somewhat unfamiliar with how to use math notation online
$G = <x_1, ..., x_n|(relations go here)>$ ,
$H = <y_1, ..., y_m|(relations go here)>$
$\Omega_1 = \{x_1, ..., x_n\}$
$\Omega_2 = \{y_1, ..., y_m\}$
$\forall x_i \in \Omega_1  \forall y_j \in \Omega_2  1\leq i\leq n, 1\leq j\leq m,   (x_iy_jx_i^{-1} \in H \wedge x_i^{-1}y_jx_i \in H )\implies H \trianglelefteq G$
I hope that formalizes the statement to which I am seeking a proof of.
 A: Let $G$ be a group and $H$ a subgroup with generating sets $S$ and $T$, respectively.  The statement you want to prove can be broken up into $2$ parts:
1) If $gtg^{-1} \in H$ for all $g \in G$ and all generators $t\in T$, then $H \trianglelefteq G$.
2) If $shs^{-1} \in H$ for all $h\in H$ and all generators $s \in S$, then $H \trianglelefteq G$.
I'll prove the first statement; see if you can work out the second and then combine the two.
Assume that $gtg^{-1} \in H$ for all $t \in T$ and $g \in G$.  We will show that $H \trianglelefteq G$.  We first show that $gt^{-1} g^{-1} \in H$ for all $t \in T$ and $g \in G$.  Given such $t$ and $g$, then by assumption $g^{-1} t g = g^{-1} t (g^{-1})^{-1} = h \in H$ for some $h \in H$.  Since $H$ is closed under inverses, then 
$$
g t^{-1} g^{-1} = (g^{-1} t g)^{-1} = h^{-1} \in H \, .
$$
Thus $g t^e g^{-1} \in H$ for all $t \in T$, $g \in G$, and $e \in \{\pm 1\}$.
We now show $H \trianglelefteq G$.  The key trick is that $g(xy)g^{-1} = (gxg^{-1})(gyg^{-1})$.  Since $T$ generates $H$, given $h \in H$ then $h = t_1^{e_1} \cdots t_n^{e_n}$ for some $t_1, \ldots, t_n \in T$ and $e_1, \ldots, e_n \in \{\pm 1\}$.  Then
\begin{align*}
ghg^{-1} &= g(t_1^{e_1} \cdots t_n^{e_n}) g^{-1} = gt_1^{e_1} g^{-1} gt_2 g^{-1} \cdots g t_n^{e_n} g^{-1} \, .
\end{align*}
Since $gt_i^{e_i}g^{-1} \in H$ for each $i$ by assumption and the first result we proved, then so is their product, hence $ghg^{-1} \in H$ and $H$ is normal.
Do you see how to prove step 2) in a similar manner?
